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Nonparametric Analysis
In the last
several issues of Reliability HotWire, we have looked at
various ways of performing parameter estimations on life data sets.
Nonparametric analysis allows the analyst to characterize life data
without assuming an underlying distribution. This can be advantageous
because it avoids some of the dangers inherent in an analysis where an
inappropriate distribution is assumed. However, the analysis is limited to
reliability estimates only for the failure times in the data set, thus
making it impossible to make reliability predictions outside the range of
data values. While this somewhat limits the usefulness of this method,
nonparametric analysis can be used as a benchmark against which other
parametric analyses can be compared for accuracy. ReliaSoft's Weibull++
software provides a nonparametric analysis utility.
There are several
methods for conducting a nonparametric analysis, including the
Kaplan-Meier, simple actuarial and standard actuarial methods. A method
for attaching confidence bounds to the results of these nonparametric
analysis techniques can also be developed. The basis of nonparametric life
data analysis is the empirical cdf function, which is given by:
Kaplan-Meier
Estimator
The Kaplan-Meier
estimator, also known as the product limit estimator, can be used to
calculate values for nonparametric reliability for data sets with multiple
failures and suspensions. The equation of the estimator is given by:
where m is the
total number of data points and n is the total number of units
under analysis. The variable ni is defined by:
where rj
is the number of failures in the jth data group, and sj
is the number of suspensions in the jth data group. Note that the
reliability estimate is only calculated for times at which one or more
failures occurred. For the sake of calculating the value of ni
at time values that have failures and suspensions, it is assumed that the
suspensions occur slightly after the failures, so that the suspended units
are considered to be operating and included in the count of ni.
Kaplan-Meier
Estimator Example
A group of 20 units are
put on a life test with the following results.
| Number in State |
State (F or S) |
State End Time |
| 3 |
F |
9 |
| 1 |
S |
9 |
| 1 |
F |
11 |
| 1 |
S |
12 |
| 1 |
F |
13 |
| 1 |
S |
13 |
| 1 |
S |
15 |
| 1 |
F |
17 |
| 1 |
F |
21 |
| 1 |
S |
22 |
| 1 |
S |
24 |
| 1 |
S |
26 |
| 1 |
F |
28 |
| 1 |
F |
30 |
| 1 |
S |
32 |
| 2 |
S |
35 |
| 1 |
S |
39 |
| 1 |
S |
41 |
Using the data and the
Kaplan-Meier estimator, the following table of reliability estimates can
be constructed.
| State End Time |
Number of Failures, ri |
Number of Suspensions, si |
Available Units, ni |
|
 |
| 9 |
3 |
1 |
20 |
0.850 |
0.850 |
| 11 |
1 |
0 |
16 |
0.938 |
0.797 |
| 12 |
0 |
1 |
15 |
1.000 |
0.797 |
| 13 |
1 |
1 |
14 |
0.929 |
0.740 |
| 15 |
0 |
1 |
12 |
1.000 |
0.740 |
| 17 |
1 |
0 |
11 |
0.909 |
0.673 |
| 21 |
1 |
0 |
10 |
0.900 |
0.605 |
| 22 |
0 |
1 |
9 |
1.000 |
0.605 |
| 24 |
0 |
1 |
8 |
1.000 |
0.605 |
| 26 |
0 |
1 |
7 |
1.000 |
0.605 |
| 28 |
1 |
0 |
6 |
0.833 |
0.505 |
| 30 |
1 |
0 |
5 |
0.800 |
0.404 |
| 32 |
0 |
1 |
4 |
1.000 |
0.404 |
| 35 |
0 |
1 |
3 |
1.000 |
0.404 |
| 39 |
0 |
1 |
2 |
1.000 |
0.404 |
| 41 |
0 |
1 |
1 |
1.000 |
0.404 |
As can be
determined from the preceding table, the reliability estimates for the
failure times are:
| Failure Time |
Reliability Estimate |
| 9 |
85.0% |
| 11 |
79.7% |
| 13 |
74.0% |
| 17 |
67.3% |
| 21 |
60.5% |
| 28 |
50.5% |
| 30 |
40.4% |
Simple Actuarial
Method
The simple actuarial
method is an easy-to-use form of nonparametric data analysis that can be
used for multiply censored data that are arranged in intervals. This
method is based on calculating the number of failures in a time interval, rj,
versus the number of operating units in that time period, ni.
The equation for the reliability estimator for the simple actuarial
method is given by:
where m is the
total number of intervals and n is the total number of units under
analysis. The variable ni is defined by:
where rj
is the number of failures in the jth data group and sj
is the number of suspensions in the jth data group.
Simple Actuarial
Method Example
A group of 55 units are put on a life test during which the units are evaluated every 50 hours. The test had the following results:
| Start Time |
End Time |
Number of Failures, ri |
Number of Suspensions, si |
| 0 |
50 |
2 |
4 |
| 50 |
100 |
0 |
5 |
| 100 |
150 |
2 |
2 |
| 150 |
200 |
3 |
5 |
| 200 |
250 |
2 |
1 |
| 250 |
300 |
1 |
2 |
| 300 |
350 |
2 |
1 |
| 350 |
400 |
3 |
3 |
| 400 |
450 |
3 |
4 |
| 450 |
500 |
1 |
2 |
| 500 |
550 |
2 |
1 |
| 550 |
600 |
1 |
0 |
| 600 |
650 |
2 |
1 |
The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in
the calculation of the reliability estimates for
the simple actuarial equation.
| Start Time |
End Time |
Number of Failures, ri |
Number of Suspensions, si |
Available Units, ni |
|
|
| 0 |
50 |
2 |
4 |
55 |
0.964 |
0.964 |
| 50 |
100 |
0 |
5 |
49 |
1.000 |
0.964 |
| 100 |
150 |
2 |
2 |
44 |
0.955 |
0.920 |
| 150 |
200 |
3 |
5 |
40 |
0.925 |
0.851 |
| 200 |
250 |
2 |
1 |
32 |
0.938 |
0.798 |
| 250 |
300 |
1 |
2 |
29 |
0.966 |
0.770 |
| 300 |
350 |
2 |
1 |
26 |
0.923 |
0.711 |
| 350 |
400 |
3 |
3 |
23 |
0.870 |
0.618 |
| 400 |
450 |
3 |
4 |
17 |
0.824 |
0.509 |
| 450 |
500 |
1 |
2 |
10 |
0.900 |
0.458 |
| 500 |
550 |
2 |
1 |
7 |
0.714 |
0.327 |
| 550 |
600 |
1 |
0 |
4 |
0.750 |
0.245 |
| 600 |
650 |
2 |
1 |
3 |
0.333 |
0.082 |
As can be determined
from the preceding table, the reliability estimates for the failure times
are:
Failure Period
End Time |
Reliability
Estimate |
| 50 |
96.4% |
| 150 |
92.0% |
| 200 |
85.1% |
| 250 |
79.8% |
| 300 |
77.0% |
| 350 |
71.1% |
| 400 |
61.8% |
| 450 |
50.9% |
| 500 |
45.8% |
| 550 |
32.7% |
| 600 |
24.5% |
| 650 |
8.2% |
Standard Actuarial
Method
The standard actuarial
model is a variation of the simple actuarial model that involves adjusting
the value for the number of operating units in an interval. The
Kaplan-Meier and simple actuarial methods assume that the suspensions in a
time period or interval occur at the end of that interval, after the
failures have occurred. The standard actuarial model assumes that the
suspensions occur in the middle of the interval, which has the effect of
reducing the number of available units in the interval by half of the
suspensions in that interval or:
With this adjustment, the calculations are carried out just as they were for the simple actuarial model,
or:
Standard Actuarial
Method Example
In this example, we
will apply the standard actuarial method to the data set used in the
simple actuarial method example. The solution to this example is similar to that of
the previous example, with the exception of the inclusion of the ni'
term, which is used in the preceding equation. Applying this equation to the data, we can generate the following table:
| Start Time |
End Time |
Number of Failures, ri |
Number of Suspensions, si |
Adjusted Units, ni' |
|
|
| 0 |
50 |
2 |
4 |
53 |
0.962 |
0.962 |
| 50 |
100 |
0 |
5 |
46.5 |
1.000 |
0.962 |
| 100 |
150 |
2 |
2 |
43 |
0.953 |
0.918 |
| 150 |
200 |
3 |
5 |
37.5 |
0.920 |
0.844 |
| 200 |
250 |
2 |
1 |
31.5 |
0.937 |
0.791 |
| 250 |
300 |
1 |
2 |
28 |
0.964 |
0.762 |
| 300 |
350 |
2 |
1 |
25.5 |
0.922 |
0.702 |
| 350 |
400 |
3 |
3 |
21.5 |
0.860 |
0.604 |
| 400 |
450 |
3 |
4 |
15 |
0.800 |
0.484 |
| 450 |
500 |
1 |
2 |
9 |
0.889 |
0.430 |
| 500 |
550 |
2 |
1 |
6.5 |
0.692 |
0.298 |
| 550 |
600 |
1 |
0 |
4 |
0.750 |
0.223 |
| 600 |
650 |
2 |
1 |
2.5 |
0.200 |
0.045 |
As can be determined
from the preceding table, the reliability estimates for the failure times
are:
Failure Period
End Time |
Reliability
Estimate |
| 50 |
96.2% |
| 150 |
91.8% |
| 200 |
84.4% |
| 250 |
79.1% |
| 300 |
76.2% |
| 350 |
70.2% |
| 400 |
60.4% |
| 450 |
48.4% |
| 500 |
43.0% |
| 550 |
29.8% |
| 600 |
22.3% |
| 650 |
4.5% |
Nonparametric
Confidence Bounds
Confidence bounds for
nonparametric reliability can be calculated in a method similar to that of
parametric confidence bounds. The difficulty in dealing with nonparametric
data lies in the estimation of the variance. To estimate the for
nonparametric data, the Weibull++
software uses Greenwood's formula:
where m is the
total number of intervals and n is the total number of units under
analysis. The variable ni is defined by:
where rj
is the number of failures in the jth data group and sj
is the number of suspensions in the jth data group. Once the variance has been calculated, the standard error can be determined by taking the square root of the variance,
This information can then be applied to determine the confidence bounds,
where:
and a
is the desired confidence level for the 1-sided confidence bounds.
Nonparametric
Confidence Bounds Example
In this example, we
will determine the 1-sided confidence bounds for the reliability estimates in
the standard actuarial method example above, with a 97.5% confidence level. Once again, this type of problem is most readily solved by constructing a table similar to the following:
| Failure Time |
Reliability Estimate |
Number of Failures |
Adjusted Units |
ri/ni' |
Variance |
Error |
w |
Lower CL |
Upper CL |
| 50 |
0.962 |
2 |
53 |
0.0377 |
0.0007 |
0.0262 |
4.108 |
0.861 |
0.991 |
| 150 |
0.918 |
2 |
43 |
0.0465 |
0.0016 |
0.0397 |
2.797 |
0.799 |
0.969 |
| 200 |
0.844 |
3 |
37.5 |
0.0800 |
0.0030 |
0.0547 |
2.257 |
0.706 |
0.924 |
| 250 |
0.791 |
2 |
31.5 |
0.0635 |
0.0040 |
0.0630 |
2.107 |
0.642 |
0.888 |
| 300 |
0.762 |
1 |
28 |
0.0357 |
0.0045 |
0.0668 |
2.059 |
0.609 |
0.868 |
| 350 |
0.702 |
2 |
25.5 |
0.0784 |
0.0054 |
0.0737 |
1.996 |
0.542 |
0.825 |
| 400 |
0.604 |
3 |
21.5 |
0.1395 |
0.0068 |
0.0823 |
1.964 |
0.438 |
0.750 |
| 450 |
0.484 |
3 |
15 |
0.2000 |
0.0082 |
0.0907 |
2.038 |
0.315 |
0.656 |
| 500 |
0.430 |
1 |
9 |
0.1111 |
0.0091 |
0.0953 |
2.142 |
0.260 |
0.618 |
| 550 |
0.298 |
2 |
6.5 |
0.3077 |
0.0104 |
0.1020 |
2.602 |
0.140 |
0.524 |
| 600 |
0.223 |
1 |
4 |
0.2500 |
0.0100 |
0.1000 |
3.098 |
0.085 |
0.471 |
| 650 |
0.045 |
2 |
2.5 |
0.8000 |
0.0036 |
0.0599 |
15.69 |
0.003 |
0.423 |
The following plot
illustrates these results graphically:
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